Question

use conservation of energy equation

Answer 1

The conservation of energy principle states that the total energy in a closed system remains constant over time. In other words, energy cannot be created or destroyed; it can only be transformed from one form to another. The conservation of energy equation often varies depending on the context, but a common form for mechanical energy is represented as:

\[
E_{\text{initial}} = E_{\text{final}}
\]

Where:
- \(E_{\text{initial}}\) is the total initial energy of the system,
- \(E_{\text{final}}\) is the total final energy of the system.

For example, in a mechanical system involving potential energy (\(PE\)) and kinetic energy (\(KE\)), the equation might look like this:

\[
PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}}
\]

Where:
- \(PE\) is the potential energy, which can be expressed as \(PE = mgh\) (mass × gravitational acceleration × height),
- \(KE\) is the kinetic energy, expressed as \(KE = \frac{1}{2}mv^2\) (one-half of mass times velocity squared).

### Example Problem

Suppose a ball of mass \(m = 2\, \text{kg}\) is dropped from a height of \(h = 10\, \text{m}\). We want to determine its speed just before it hits the ground.

1. **Calculate Initial Energy**:
- Initial Potential Energy \(PE_{\text{initial}} = mgh = 2\, \text{kg} \times 9.81\, \text{m/s}^2 \times 10\, \text{m} = 196.2\, \text{J}\)
- Initial Kinetic Energy \(KE_{\text{initial}} = 0\, \text{J}\) (the ball is at rest before being dropped)

Thus,
\[
E_{\text{initial}} = PE_{\text{initial}} + KE_{\text{initial}} = 196.2\, \text{J} + 0\, \text{J} = 196.2\, \text{J}
\]

2. **Calculate Final Energy** (just before hitting the ground):
- Final Potential Energy \(PE_{\text{final}} = 0\, \text{J}\) (at ground level)
- Let the final kinetic energy be \(KE_{\text{final}} = \frac{1}{2}mv^2\)

Using conservation of energy:
\[
E_{\text{initial}} = E_{\text{final}}
\]
\[
196.2\, \text{J} = 0\, \text{J} + KE_{\text{final}}
\]
Thus,
\[
KE_{\text{final}} = 196.2\, \text{J}
\]

Substituting \(KE_{\text{final}} = \frac{1}{2}mv^2\):
\[
196.2\, \text{J} = \frac{1}{2} \times 2\, \text{kg} \times v^2
\]
\[
196.2\, \text{J} = 1\, \text{kg} \times v^2
\]
\[
v^2 = 196.2
\]
\[
v = \sqrt{196.2} \approx 14.0\, \text{m/s}
\]

Thus, the speed of the ball just before it hits the ground is approximately \(14.0\, \text{m/s}\).